Problem: Solve for $x$ and $y$ using elimination. ${2x-y = 0}$ ${3x+y = 15}$
Answer: We can eliminate $y$ by adding the equations together when the $y$ coefficients have opposite signs. Add the equations together. Notice that the terms $-y$ and $y$ cancel out. $5x = 15$ $\dfrac{5x}{{5}} = \dfrac{15}{{5}}$ ${x = 3}$ Now that you know ${x = 3}$ , plug it back into $\thinspace {2x-y = 0}\thinspace$ to find $y$ ${2}{(3)}{ - y = 0}$ $6-y = 0$ $6{-6} - y = 0{-6}$ $-y = -6$ $\dfrac{-y}{{-1}} = \dfrac{-6}{{-1}}$ ${y = 6}$ You can also plug ${x = 3}$ into $\thinspace {3x+y = 15}\thinspace$ and get the same answer for $y$ : ${3}{(3)}{ + y = 15}$ ${y = 6}$